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From: panos@utcsri.UUCP (Panos Economopoulos)
Newsgroups: net.puzzle
Subject: Re: More on Interview Questions...derivative of x!
Message-ID: <926@utcsri.UUCP>
Date: Mon, 25-Mar-85 13:58:07 EST
Article-I.D.: utcsri.926
Posted: Mon Mar 25 13:58:07 1985
Date-Received: Mon, 25-Mar-85 14:21:29 EST
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Mark Brader correctly pointed out an elementary error in my
calculations of the derivative of the Gamma function.
It seems that the resulting integral is not as easy to evaluate
as it initially (erroneously) appeared

However,

1. Of course, x! is generally defined as x(x-1)(x-2)..2*1 for 
  non-negative integers x  (0!=1). In that case it is a discrete
  variable function, and the only thing resembling a derivative
  is a difference function D(x) = [ f(x+1) - f(x) ] / 1
  which in our case is  (x+1)! -x! = x * x!

2. If, however, we consider the values x! to be just the values of
  the Gamma function at the integral values of the variable x, then
  it is of interest to derive the derivative (which doesn't seem trivial).
  In this case, the values of the derivative G' at the integral values
  of x, will give us the slope of the Gamma at these points.
  This will be smaller than the corresponding D(x) found above, because
  in this case D(x) represents the slopes of the  linear cords joining
  the points (x,x!) on the plot of the Gamma function and the Gamma
  is a concave function.

3. Looking up a Schaum's Mathematical Handbook of formulas and tables,
  I found that the derivative of Gamma is given by the following formula

  G'(x) / G(x) =

	-g + ( 1/1 - 1/x ) +

	( 1/2 - 1/(x+1) ) +

	.... +

	( 1/n - 1/(x+n-1) ) + 

	.....

  where g is Euler's constant.  For x=1,  G'(1) = -g.


I think I'll stick with    x * x! :-)
-- 

					Panos Economopoulos

UUCP:   {decvax,linus,ihnp4,uw-beaver,allegra,utzoo}!utcsri!panos
CSNET:  panos@toronto